I'm interested in finding the curve $q(t):[0,1] \rightarrow \mathbb{R}^+$ that satisfies the boundary conditions $q(0)=q(1)=0$, the integral condition $\int_0^1q(t)dt=a>0$, and that minimizes the functional $J(q,\dot{q}, t)=\int_0^1\sqrt{1+\dot{q}^2}dt$, that is the length of the curve $q$ ($\dot{q}$ indicates the first derivative of $q$). Other conditions that might be relevant are: $q \in C^n$ for some $n>2$, $q$ is strictly concave with maximum at $1/2$. Practically, something that looks like a concave parabola, a catenary or a hyperbole.
Applying Hamilton's variational principle without the conditions to this simple functional $J$ obviously returns a linear curve, so I'm looking for a way of including the conditions (especially the integral one) and still apply Hamilton's principle. My first idea was that if $q$ is stationary for $J$, it will be stationary also for $J+\lambda a=\int_0^1\sqrt{1+\dot{q}^2}dt+\lambda\int_0^1 q(t)dt$ for some $\lambda\in\mathbb{R}$, so one might instead try the functional $I=\int_0^1\sqrt{1+\dot{q}^2}+\lambda q(t)dt$, with respective Lagrangian $L=\sqrt{1+\dot{q}^2}+\lambda q$. Applying Hamilton's principle to this functional returns the curve $q(t)=\sqrt{1-(\lambda t)^2}-1$, which is not what i'm looking for.
I'm looking for ways of including both the integral and boundary conditions in the application of Hamilton's principle, hopefully by simple playing around with the functional. The problem arose after looking at a plastic peel covering a bowl with some rising pizza dough. The dough produces gassy volume that expands the peel more or less like a balloon. Thus the gas produced occupies a certain volume (area $a$ in the 2 dimensional case), and we want to minimize the length of the curve of the plastic peel to minimize the amount of stretching required.
I think you are almost all correct, except for one step which you missed an arbitrary integration constant.
After applying the Euler Lagrange equation $$ \frac{d}{dt}\frac{\dot q}{\sqrt{1+\dot q^2}} = \lambda $$ Here you miss a constant term when integrating $t$ $$ \frac{\dot q}{\sqrt{1+\dot q^2}} = \lambda t+c\\ \dot q^2=\frac{(\lambda t+c)^2}{1-(\lambda t+c)^2}\\ \dot q=\pm\sqrt\frac{(\lambda t+c)^2}{1-(\lambda t+c)^2} $$ Doing indefinite integration of $q$ gets $$ q(t)=\pm\frac{1}{\lambda}\sqrt{1-(\lambda t+c)^2} +K $$ We can pick the $+$ solution using the boundary condition $q(0)=q(1)=0$, by symmetry we can see $$ c^2=(\lambda+c)^2\\ c=-\frac{\lambda}{2} $$ Thus $$ q(t)=\frac{1}{\lambda}\sqrt{1-\lambda^2 (t-\frac12)^2} -\frac{1}{\lambda}\sqrt{1- \frac{\lambda^2}{4}} $$ This is an equation for an arc of a circle. $$ (q(t)-K)^2 + (t-\frac 12)^2=\frac{1}\lambda^2{} $$